At a particular instant, a stationary observer on the ground sees a package falling with a speed v1 at an angle θ to the vertical. To a pilot flying horizontally with a constant speed v relative to the ground, the package appears to be falling vertically with a speed v2 ( at that same instant ). The speed of the pilot relative to the ground (.v.) is

Question

At a particular instant, a stationary observer on the ground sees a package falling with a speed v1 at an angle θ to the vertical. To a pilot flying horizontally with a constant speed v relative to the ground, the package appears to be falling vertically with a speed v2 ( at that same instant ). The speed of the pilot relative to the ground (.v.) is (A) ( textv12 + textv22)1 / 2 (B) ( mathrmv1- mathrmv2)( mathrmv2- mathrmv1) (C) ( textv12 - textv22)1 / 2 (D)  textv1 - textv2

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-4cphzldbq3f5.png" /> we can see from the diagram, vector v1 is resultant of two vectors v and v2 Hence v2+v22=v12 ⇒ v=√ textv12 - textv22

$(v_{1}^{2} + v_{2}^{2})^{1/2}$

$(v_{1} - v_{2}) (v_{2} - v_{1})$

$(v_{1}^{2} - v_{2}^{2})^{1/2}$

$v_{1} - v_{2}$

we can see from the diagram, vector $v_{1}$ is resultant of two vectors $v$ and $v_{2}$
Hence $v^{2} + v_{2}^{2} = v_{1}^{2}$
$\Rightarrow v = v_{1}^{2} - v_{2}^{2}$

(v12​+v22​)1/2

(v1​−v2​)(v2​−v1​)

(v12​−v22​)1/2

v1​−v2​

we can see from the diagram, vector v1​ is resultant of two vectors v and v2​ Hence v2+v22​=v12​ ⇒v=v12​−v22​​

$(v_{1}^{2} + v_{2}^{2})^{1/2}$

$(v_{1} - v_{2}) (v_{2} - v_{1})$

$(v_{1}^{2} - v_{2}^{2})^{1/2}$

$v_{1} - v_{2}$

we can see from the diagram, vector $v_{1}$ is resultant of two vectors $v$ and $v_{2}$
Hence $v^{2} + v_{2}^{2} = v_{1}^{2}$
$\Rightarrow v = v_{1}^{2} - v_{2}^{2}$